Question Number 128845 by Dwaipayan Shikari last updated on 10/Jan/21
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}.\mathrm{4}}{\left(\mathrm{5}.\mathrm{1}!\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{1}.\mathrm{4}.\mathrm{6}.\mathrm{9}}{\left(\mathrm{5}^{\mathrm{2}} .\mathrm{2}!\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{1}.\mathrm{4}.\mathrm{6}.\mathrm{9}.\mathrm{11}.\mathrm{14}}{\left(\mathrm{5}^{\mathrm{3}} .\mathrm{3}!\right)^{\mathrm{2}} }+…=\frac{{b}^{\mathrm{2}} \sqrt{\frac{{b}−\sqrt{{b}}}{\mathrm{2}}}}{{a}\pi} \\ $$$${Find}\:\mathrm{5}{a}−\mathrm{8}{b} \\ $$
Answered by mindispower last updated on 10/Jan/21
![=1+Σ_(n≥1) ((Π_(k=0) ^(n−1) (1+5k)(4+5k))/((n+1).(5^n .n!)^2 )) =1+Σ_(n≥1) ((5^(2n) Π_(k=0) ^(n−1) (k+(1/5)).Π_(k=0) ^(n−1) (k+(4/5)))/(5^(2n) .(n+1)!.n!)) =1+Σ_(n≥1) ((((1/5))_n .((4/5))_n )/((2)_n )).(((1)^n )/(n!)) =_2 F_1 ((1/5),(4/5);2;[1])=((Γ(2−1)Γ(2))/(Γ(2−(1/5))Γ(2−(4/5))))=(1/(Γ((9/5))Γ((6/5)))) =Γ((9/5))=(4/5)Γ((4/5)) Γ((6/5))=(1/5)Γ((1/5)) S=((25)/(4Γ((1/5))Γ((4/5))))=((25)/(4π))sin((π/5))=((25)/(4π)).(1/4).(√(10−2(√5))) =((25)/(8π)).(√((5−(√5))/2)),b=5,a=8 5a−8b=0](https://www.tinkutara.com/question/Q128860.png)
$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{1}+\mathrm{5}{k}\right)\left(\mathrm{4}+\mathrm{5}{k}\right)}{\left({n}+\mathrm{1}\right).\left(\mathrm{5}^{{n}} .{n}!\right)^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{5}^{\mathrm{2}{n}} \underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{5}}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{4}}{\mathrm{5}}\right)}{\mathrm{5}^{\mathrm{2}{n}} .\left({n}+\mathrm{1}\right)!.{n}!} \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{5}}\right)_{{n}} .\left(\frac{\mathrm{4}}{\mathrm{5}}\right)_{{n}} }{\left(\mathrm{2}\right)_{{n}} }.\frac{\left(\mathrm{1}\right)^{{n}} }{{n}!} \\ $$$$=_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}},\frac{\mathrm{4}}{\mathrm{5}};\mathrm{2};\left[\mathrm{1}\right]\right)=\frac{\Gamma\left(\mathrm{2}−\mathrm{1}\right)\Gamma\left(\mathrm{2}\right)}{\Gamma\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\mathrm{2}−\frac{\mathrm{4}}{\mathrm{5}}\right)}=\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{9}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{6}}{\mathrm{5}}\right)} \\ $$$$=\Gamma\left(\frac{\mathrm{9}}{\mathrm{5}}\right)=\frac{\mathrm{4}}{\mathrm{5}}\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\Gamma\left(\frac{\mathrm{6}}{\mathrm{5}}\right)=\frac{\mathrm{1}}{\mathrm{5}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$${S}=\frac{\mathrm{25}}{\mathrm{4}\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right)}=\frac{\mathrm{25}}{\mathrm{4}\pi}{sin}\left(\frac{\pi}{\mathrm{5}}\right)=\frac{\mathrm{25}}{\mathrm{4}\pi}.\frac{\mathrm{1}}{\mathrm{4}}.\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$=\frac{\mathrm{25}}{\mathrm{8}\pi}.\sqrt{\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}}},{b}=\mathrm{5},{a}=\mathrm{8} \\ $$$$\mathrm{5}{a}−\mathrm{8}{b}=\mathrm{0} \\ $$$$ \\ $$